SATHEE: Rotational Motion Question 129

Rotational Motion Question 129

Question: A thick walled hollow sphere has outer radius $R$ . It rolls down an inclined plane without slipping and its speed at the bottom is $v$ . If the inclined plane is frictionless and the sphere slides down without rolling, its speed at the bottom is $5 v / 4$ . What is the radius of gyration of the sphere?

Options:

A) $\frac{R}{\sqrt{2}}$

B) $\frac{R}{2}$

C) $\frac{3 R}{4}$

D) $\frac{\sqrt{3} R}{4}$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Case I: $m g h = \frac{1}{2} m v^{2} + \frac{1}{2} m k^{2} ω^{2}$

$C a s e I I :$

$m g h = \frac{1}{2} m \times {(\frac{5 v}{4})}^{2}$

$\frac{1}{2} m k^{2} \frac{v^{2}}{R^{2}} + \frac{1}{2} m v^{2} = \frac{1}{2} m \times \frac{25 v^{2}}{16}$

$\frac{k^{2}}{R^{2}} + 1 = \frac{25}{16} \Rightarrow k = \frac{3 R}{4}$

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Rotational Motion Question 129

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